Question

given y=tanxy+x^2, find dy/dx

Answer 1

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Answer 2

y' = sec^2 (xy)* [ y + xdy/dx) + 2x

Answer 3

thank you laplace2!

Answer 4

y' = sec^2 ( xy)*y + sec^2 (xY) * x *y' + 2x , distributing

Answer 5

collect like terms

Answer 6

y ' [ 1 - sec^(xy) * x ] = sec^2 (xy)*y + 2x

Answer 7

y' = ( sec^2 (xy)*y + 2x ) / ( 1 - sec^2(xy)*x)